Asymptotic Bode Plot & Lead-Lag Compensator

Transcription

1 Asymptotic Bode Plot & Lead-Lag Compensator. Introduction Consider a general transfer function Ang Man Shun G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer function G( is very useful in frequency response analysis. As now the transfer function, or using another name, the gain is a function of frequency, it means that amplification ability of such system change as change, where the is the input or operating frequency. There are some methods to investigate the behaviour of such amplification ability changes : Bode Plot, Nyquist Plot, Nichol Plot. The Bode P lot consists of 2 plots : the magnitude plot and the phase plot, it is an asymptotic plot, an approximation plot. The magnitude Bode Plot is 20log G( (y-axis vs (x-axis, log scale The magnitude axis is expressed in db of powers, it is a convention. The frequency axis is expressed in log scale, it is also a convention, with a very good reason. 2. Review of some related mathematics Log log AB = log A + log B log C log D CD log AB CD = log A + log B log C log D ( Standard Form log + a = log a a + = log a + log Approximation Consider log, a is called the corner frequency when a a a a = a = a log 0 log log a 20 log = log = +20 log a Flat horizontal line Stright line increasing with 20dB per decade Arc-Tan tan = ± π 2 tan 0 = 0

2 3. Example For example, consider a degree transfer function with 5 zeros and poles. G(s = Where a, b are zeros, c, d, e are poles. A (s + a (s + b (s + c (s + d (s + e Before doing the Bode Plot analysis, turn the transfer function into Standard Form G(s = Aab cde a + b + k = a + b + k = Aab c + d + e + c + d + e + cde The magnitude (in db of powers is thus k 20 log G(s = 20 log a + b + c + d + e + = 20 log k + 20 log s a + s + 20 log b + s 20 log c + s 20 log d + s 20 log e + The Bode Plot is thus 4. For more general case The Effect of DC term : a horizontal line The Effect of a pole at α : For < α, no effect (since it is in log scale!, for > α, it goes down at a rate of 20dB per decade. The Effect of a zero at β : For < β, no effect (since it is in log scale!, for > β, it goes up at a rate of 20dB per decade. 2

3 More General For a simple, st order pole at α : For < α, no effect, for > α, it goes down at a rate of 20dB per decade. For a n th order pole at β : For < α, no effect, for > α, it goes down at a rate of n20 db per decade. In same logic, a m th order zero at γ will give a increasing straight with slope of m20db per decade, starting at = γ 5. Summary of Magnitude Plot The algorithm to draw Magnitude Bode Plot is Turn the transfer function into standard form (s + a a a + Find all the corner frequency For zeros, the lines go up. For poles, the lines go down. The slope of the line is 20dB the degree of the pole/zero. 6. Phase Plot Consider ( + a, the angle of this term is tan a When a ( The axis is in scale of decade, i.e. 0 time = 0.a = 0.0a tan a = tan (0. = 5.7 o tan a = tan (0.0 = 0.57 o When a ( The axis is a log scale, so consider a decade, i.e. 0 time = 0a tan a = tan (0 = 84.3 o = 00a tan a = tan (00 = 89.4 o So in general, tan 0 o a a 90 o a 3

4 7. Examples G(s = K 0 T 0 s + Let K = 4, T 0 = log 4 = 27 Corner Frequenct in magnitude plot : 0.5 Corner Frequency in phase plot : 0.05 and 5 K 0, T 0 > 0 20 log G( = 20 log K 0 20 log + /T 0 ( G( = K }{{} 0 + /T G(s = 2 s + 4

5 8. Example 2 G(s = First change it into standard form Let K 0 = 8, a =, b = 3 20 log 8 9 = K 0 (s + a (s + b 2 K 0, a, b > 0 G(s = K 0 ab 2 a + b log G( = 20 log K 0 ab 2 20 log a log b + ( ( ( K0 G( = ab }{{ 2 a + 2 b + } 0 Corner Frequency in magnitude plot :, 3 Corner Frequency in phase plot : 0. & 0, 0.3 & 30 5

6 9. Lead Lag Compensator + T s + αt s Purpose of Compensator System is not always perfect naturally, some improvements can be made by adding a special type of controller, the Lead-Lag Compensator The improvements can be Increase Gain Reduce steady state Error Force he system response faster Force the system become less oscillatory Increase bandwidth Improve stability margin The Lead and the Lag Compensator There are two form, both are the same Consider the second one + αt s + T s or + T s + αt s K DC + T s + αt s K DC : DC gain, when G(s s=0 The Transfer Function of the compensator has zero and poles : T The α is key parameter here. and αt 6

7 Consider their Bode Plot Standard Form + T s K(s = K DC + αt s s + /T K(s = K DC + s /αt 20 log K( = 20 log K DC + 20 log + j K( = K }{{ DC} 0 Corner frequency of magnitude plot : Corner frequency of phase plot : Denote T = a αt = b /T 20 log + j ( + + j ( + j /T /αt T, αt 0T & 0 T and 0αT & 0 αt The following Bode Plot ignore the DC part (Assume K DC = /αt 7

8 Lead-Lag compensator K(s = KDC + Ts + Ts + α T s + α2 T s 20 log K = 20 log KDC log + j 20 log + j 20 log + j /T /α T /α2 T ( ( ( K = 2 + j + j + j /T /α T /α2 T Corner frequency in magnitude plot : ( denote as a, denote as f, denote as g T α T α2 T g<a<f Corner frequency in phase plot : 0.a, 0a, 0.f, 0f, 0.g, 0g EN D 8